About a Brooks-type theorem for improper colouring

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1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 43 (2009), Pages About a Brooks-type theorem for improper colouring Ricardo Corrêa Federal University of Ceará Campus do Pici, Bloco Fortaleza, CE Brazil correa@lia.ufc.br Frédéric Havet Mascotte, I3S (CNRS-UNSA) INRIA Route des Lucioles, BP Sophia Antipolis Cedex France fhavet@sophia.inria.fr Jean-Sébastien Sereni CNRS (LIAFA), Paris France and Institute for Theoretical Computer Science ITI and Department of Applied Mathematics KAM Charles University, Prague 1 Czech Republic sereni@kam.mff.cuni.cz Abstract Agraphisk-improperly l-colourable if its vertices can be partitioned into l parts such that each part induces a subgraph of maximum degree at most k. AresultofLovász states that for any graph G, such a partition exists if l. When k = 0, this bound can be reduced by Δ(G)+1 k+1 This work was partially supported by the Égide eco-net project 16305SB. Partially supported by the Brazilian agencies CNPq and CAPES. Partially supported by European Project ist fet Aeolus. The research in this paper was carried out while this author was a doctoral research student in Mascotte. This author s work is partially supported by the European project ist fet Aeolus.

2 220 R. CORRÊA, F. HAVET AND J-S. SERENI Brooks Theorem, unless G is complete or an odd cycle. We study the following question, which can be seen as a generalisation of the celebrated Brooks Theorem to improper colouring: does there exist a polynomialtime algorithm that decides whether a graph G of maximum degree Δ has k-improper chromatic number at most Δ+1 1? We show that k+1 the answer is no, unless P = N P, when Δ = l(k +1), k 1, and l + l 2k + 3. We also show that, if G is planar, k =1ork =2, Δ=2k +2, and l = 2, then the answer is still no, unless P = N P. These results answer some questions of Cowen, Goddard and Jesurum [J. Graph Theory 24(3) (1997), ]. Introduction An l-colouring of a graph G =(V,E) is a mapping c : V {1, 2,...,l}. For any vertex v V,theimpropriety of v under c is im c (v) := {u V : uv E and c(u) =c(v)}. A colouring is k-improper provided that the impropriety of every vertex is at most k. A 0-improper colouring is proper. A graph is k-improperly l-colourable if it admits a k-improper l-colouring. The k-improper chromatic number is c k (G) :=min{l : G is k-improperly l-colourable}. In particular, c o (G) is the chromatic number χ(g) of the graph G. Since the early nineties, a lot of work has been devoted to various aspects of improper colourings, both from a purely theoretical point of view [7, 8, 14, 20, 21] and in relation with frequency assignment issues [12, 13, 16]. Let us note that improper colourings are also called in the literature defective colourings. For all integers k and l, let k-imp l-col be the following problem: INSTANCE: a graph G. QUESTION: is Gk-improperly l-colourable? Cowen et al. [8] showed that the problem k-imp l-col is N P -complete for all pairs (k,l) of integers with k 1andl 2. When l 3, this is not very surprising since it is already hard to determine whether a given graph is properly 3-colourable. On the contrary, determining if a graph is 2-colourable, i.e. bipartite, can be done in polynomial-time, whereas it is N P -complete to know if it is k-improper 2-colourable as soon as k>0. Of even more interest is the question of complexity of k-imp l-col when restricted to graphs with maximum degree (k + 1)l. Indeed, Lovász [17] proved that, for any graph G, it holds that c k (G) Δ(G)+1 k+1, where Δ(G)isthemaximumdegree of G. Whenk = 0, this is the usual bound χ(g) Δ(G) + 1. Brooks Theorem [6] states that this upper bound can be decreased by one, provided that G is neither complete nor an odd cycle, which can be checked in polynomial-time. Extensions of Brooks Theorem have also been considered. A well-known conjecture of Borodin and Kostochka [5] states that every graph of maximum degree

3 ABOUT A BROOKS-TYPE THEOREM 221 Δ 9 and chromatic number at least Δ has a Δ-clique. Reed [19] proved that this is true when Δ is sufficiently large, thus settling a conjecture of Beutelspacher and Herring [4]. Further information about this problem can be found in the monograph of Jensen and Toft [15, Problem 4.8]. Generalisation of this problem has also been studied by Farzad, Molloy, and Reed [11] and Molloy and Reed [18]. In particular, it is proved [18] that determining whether a graph with large constant maximum degree Δ is (Δ k)-colourable can be done in linear time if (k +1)(k +2) Δ. This threshold is optimal by a result of Emden-Weinert, Hougardy, and Kreuter [10], since they proved that for any two constants Δ and k Δ 3 such that (k+1)(k+2) > Δ, determining whether a graph of maximum degree Δ is (Δ k)-colourable is N P - complete. It is natural to ask whether analogous results can be found for improper colouring. This first problem to grapple with is the existence, or not, of a Brooks-like theorem for improper colouring: does there exist a polynomial-time algorithm that decides whether a graph G of maximum degree Δ has k-improper chromatic number at most Δ+1 k+1 1? Proving that k-imp l-col is N P -complete when restricted to graphs with maximum degree (k + 1)l would provide a negative answer to this question unless P = N P. Cowen et al. [8] proved that k-imp 2-COL is N P -complete for the class of graphs with maximum degree 2(k + 1), and asked what happens when l 3. In this paper, we prove that k-imp l-col restricted to graphs with maximum degree (k +1)l is N P -complete for all integers k 1andl {3,...,s}, wheres is the biggest integer such that s + s 2k + 3 (Theorem 2). An intriguing question that remains unanswered is the complexity of this problem for larger values of l. Problem 1. What is the complexity of k-imp l-col restricted to graphs with maximum degree (k +1)l when l + l>2k +3? We conjecture that is always N P -complete. As an evidence, we prove the N P - completeness when k =1andl = 4 (Theorem 6). In view of these negative results, one may ask what happens for planar graphs. It is known that every planar graph is 4-colourable [1, 3, 2], 2-improperly 3-colourable [9, 21], and Cowen et al. [8] proved that is is N P -complete to know whether a planar graph is 1-improperly 3-colourable, but without any restriction on the maximum degree. Cowen et al. [8] also proved that k-imp 2-COL is N P -complete for planar graphs, again without any restriction on the degree. In particular, they asked if 1-IMP 2- COL is still N P -complete for planar graphs with maximum degree 4 they could prove it only for maximum degree 5. In more general terms, we consider the following problem. Problem 2. What is the complexity of k-imp 2-COL restricted to planar graphs of maximum degree 2k +2? We show in Section 2 that it is NP-complete when k {1, 2}. Note that for k = 1, it settles Cowen et al. [8] question. However, we conjecture that if k is sufficiently large then k-imp 2-COL can be polynomially decided, as the answer is always affirmative.

4 222 R. CORRÊA, F. HAVET AND J-S. SERENI Conjecture 1. There exists an integer k 0 3 such that for any k k 0, any planar graph with maximum degree at most 2k +2is k-improperly 2-colourable. We end the introduction with two definitions. Given an undirected graph G, an orientation of G is any directed graph obtained from G by assigning a unique direction to each edge. Then, a vertex u is dominated by a vertex v if there is an edge between v and u directed from v to u. 1 Complexity of k-imp l-col for graphs with maximum degree (k +1)l In this section, we study the complexity of k-imp l-col restricted to graphs with maximum degree (k +1)l. The main result is the following theorem establishing the NP-completeness of the problem when k 1andl + l 2k +3. Theorem 2. Fix a positive integer k, and an integer l 3 such that l+ l 2k+3. The following problem is N P -complete: INSTANCE: a graph G with maximum degree at most (k +1)l. QUESTION: is Gk-improperly l-colourable? Remark 3. The maximum value of l is approximately 2k +4 2k +2. To prove Theorem 2, we need some preliminaries. Let k and l be two positive integers, and let H(k,l) be the graph with vertex set X Y {z} where X = (k +1)(l 1) and Y =(k + 1) such that xy is an edge unless x = z and y Y (see Figure 1). X Y z Figure 1: The complement of the graph H(3, 3). Proposition 4. The graph H(k,l) is k-improperly l-colourable, and in any k-improper l-colouring of H(k,l) the vertices of Y {z} are coloured the same. Proof. Since H(k,l)has(k+1)l+1 vertices, at least one colour class must contain at least k + 2 vertices. Observe that a vertex of X must be in a colour class containing at most k + 1 vertices since it is connected to every other vertex. Hence, the colour class with k + 2 vertices is Y {z}.

5 ABOUT A BROOKS-TYPE THEOREM 223 Lemma 5. Let G be a graph with maximum degree at most 2k +2.ThenG has an orientation D such that every vertex has indegree and outdegree at most k +1. Proof. Since every graph with maximum degree at most 2k + 2 is a subgraph of a (2k + 2)-regular graph, it suffices to prove the assertion for (2k + 2)-regular graphs. Let G be such a graph, then it admits an Eulerian cycle C. LetD be the orientation of G such that (u, v) isanarcifandonlyifu precedes v in C. Then D has indegree and outdegree at most k +1. Proof of Theorem 2. Reduction to the following problem: INSTANCE: a graph G with maximum degree at most 2k +2 l QUESTION: is Gl-colourable? Thanks to the choice of l, this problem is N P -complete by the result of Emden- Weinert et al. [10] cited in the introduction. Let G =(V,E) be a graph of maximum degree at most 2k + 2, and let D be an orientation of G with in- and outdegree at most k + 1. Such an orientation exists by Lemma 5. Let G be the graph constructed as follows: replace each vertex v of G by a copy H(v) ofh(k,l); if v dominates u in D then connect z(v) to an element of Y (u), in such a way that every vertex of Y (u) is connected to a single vertex not in H(u). The maximum degree of G is (k +1)l. Let us now prove that G is l-colourable if and only if G is k-improperly l- colourable. If G admits an l-colouring c, then for any vertex v we assign the colour c(v) to the vertices of Y (v) {z(v)} and the l 1 other colours to l 1 disjoint sets of k + 1 vertices of X(v). This yields a k-improper l-colouring of G. Conversely, suppose that G admits a k-improper l-colouring c.letcbedefined by c(v) :=c (z(v)). We prove now that c is a proper l-colouring of G: let u and v be two neighbours. Without loss of generality, v is the predecessor of u in D. Thus,the vertex z(v) is connected to an element y(u) ofy (u). Note that c (z(v)) c (z(u)), otherwise by Proposition 4, all the vertices of Y (u) Y (v) {z(u),z(v)} are coloured the same. Then y(u) would have degree k + 1 in this set which is impossible. This concludes the proof. We now extend this result to the case when k =1andl =4. Theorem 6. The following problem is N P -complete: INSTANCE: a graph G with maximum degree at most 8. QUESTION: is G 1-improperly 4-colourable? Let B be the graph with vertex set {a 1,a 2,a 3,b 1,b 2,b 3 },andxy is an edge except if there exists i {1, 2, 3} such that {x, y} = {a i,b i }. Let A be the graph with vertex set {x 1,x 2,y 1,y 2 } andwiththetwoedgesx 1 x 2 and y 1 y 2. For i {2, 3, 4}, let J i be the union of a copy A i of A and a copy B i of B, to which we add every edge xy such that (x, y) A i B i. Let A be the graph obtained from A by removing the edge y 1 y 2. We define J 1 to be the union of a copy A 1 of A and a copy B 1 of B. We let J 1 be a copy of J 1 (with A 1 and B 1 defined analogously).

6 224 R. CORRÊA, F. HAVET AND J-S. SERENI Let H := J 1 4 i=1 J i, to which we add the following edges (see Figure 2): y J1 1 yj2 1, yj1 1 yj3 1, yj1 2 yj4 1 y J 1 1 y J2 2, y J 1 1 y J3 2, y J 1 2 y J4 2 x J2 1 x J4 2, x J2 2 x J3 1, x J3 2 x J4 1 Proposition 7. The graph H is 1-improperly 4-colourable, and for any 1-improper 4-colouring of H, thesetsa i,fori {1, 2, 3, 4}, anda 1 A 1 are monochromatic. Proof. Consider a 1-improper 4-colouring of H. For every i {1, 2, 3, 4}, the colour of each vertex not belonging to A i is assigned at most twice. Therefore, all the vertices of A i must be coloured the same. The same holds also for A 1.Moreover,for every j {2, 3, 4}, the colour of the vertices of A 1 and of A j must be different from the colour of the vertices of A i for every i {1, 2, 3, 4}\{j}. Hence, A 1 and A 1 are coloured the same. Proof of Theorem 6. Reduction to the following problem, which is N P -complete [10]: INSTANCE: a graph G with degree at most 6. QUESTION: is G 4-colourable? Let G =(V,E) be a graph of maximum degree 6. By Lemma 5, let D be an orientation of G with in- and outdegree at most k +1. Let G be the graph obtained by replacing each vertex v of G by a copy H(v) of H; wesetx(v) :={x J1 1 (v),x J2 2 (v),x J 1 1 (v)} and Y (v) :={y J1 2 (v),y J 1 2 (v),x J 1 2 (v)}; if v dominates u in D, then we connect an element of Y (v) to an element of X(u) in such a way that every vertex of X(v) Y (v) is connected to a single vertex not in H(v), see Figure 2. The maximum degree of G is 8. Moreover G is 1-improper 4-colourable if and only if G is 4-colourable. Indeed, consider any proper 4-colouring of G. For every vertex v V, assign to every vertex of A 1 (v) A 1(v) the colour of v. This partial colouring of G can be extended to a 1-improper 4-colouring of G. Conversely, if C is a 1-improper 4-colouring of G, then by Proposition 7, for each v V the vertices of A 1 (v) are monochromatic. For all v V, we define C(v) to be the colour assigned to every vertex of A 1 (v). Then C is a proper 4-colouring of G: consider an edge uv of G, and say it is oriented from u to v in D. By Proposition 7, the vertex of X(u) to which the corresponding edge of G is linked has impropriety 1 in H(u). Thus, the vertex of Y (v) to which it is linked is coloured differently. Consequently, by the definition of C, we deduce that C(u) C(v), as desired. 2 Planar graphs In this section we show that 1-IMP 2-COL and 2-IMP 2-COL are N P -complete when restricted to planar graphs of bounded maximum degree.

7 ABOUT A BROOKS-TYPE THEOREM 225 A2 J2 J1 v J3 J 1 J4 H(v) Figure 2: Replacing a vertex v of G by a copy H(v) ofh. For each subgraph J i or J 1, dotted lines indicate missing edges. Theorem 8. The following problem is N P -complete: INSTANCE: a planar graph G with maximum degree 4. QUESTION: is there a 1-improper 2-colouring of G? Proof. This result is proved by slightly modifying the proof of Cowen et al. [8]. The only change is in the crossing gadget. The crossing gadget of Cowen et al. [8] has maximum degree 5, and they asked whether one with maximum degree 4 exists. We shall exhibit such a crossing gadget. We make the whole proof here for completeness. The reduction is from 3-sat. LetΦbea3-cnf. We shall construct, in polynomial time, a planar graph G Φ of maximum degree 4 such that Φ is satisfiable if and only if G Φ is 1-improperly 2-colourable. We use several gadgets with useful properties. An xy-regulator is depicted in Figure 3. There is a unique 1-improper 2-colouring of this graph, in which x, u 1, u 2,andy form a colour-class and all have impropriety zero. Note also that both x and y have degree 1 within an xy-regulator. An xyinversor, depicted in Figure 4, is a K 2,3,withxand y being any two vertices not in the same part of the bipartition. There is a unique 1-improper 2-colouring of this graph, in which x and y receive different colours, and both have impropriety zero. Note that one out of x and y has degree 2 in the inversor while the other has degree 3. The variable-gadget that represents the literals of a particular variable is constructed from the two aforementioned graphs as shown in Figure 5. We put one variable-gadget for each variable, with as many literals as needed. Each of the literals will be linked to exactly one clause. Let C 1,C 2,...,C m be the clauses of Φ. For each clause C i, we put a copy G i of the graph G. It is left to the reader to check out that the graph G, depicted in Figure 6, has the following property: in any 1-improper 2-colouring, z is coloured 1 if and only

8 226 R. CORRÊA, F. HAVET AND J-S. SERENI u 1 x y u 2 Figure 3: A regulator and its unique 1-improper 2-colouring. x y Figure 4: An inversor and its unique 1-improper 2-colouring. x x x x x x Figure 5: The vertex gadget. A double edge stands for a regulator. if at least one of p, q, r is. Then we add the edges z i z i+1,fori {1, 2,...,m 1}. z p q r Figure 6: The clause gadget. A double edge stands for a regulator. The obtained graph has maximum degree 4, and the vertex-gadgets and the clauses can be arranged so that the only edges that can cross are the ones joining vertex-gadgets to clauses. We uncross two crossing edges by using the crossing gadget CG depicted in Figure 7. The graph CG has maximum degree 4, and fulfils the following properties: (i) the graph CG is 1-improperly 2-colourable but not 1-improperly 1-colourable; (ii) in any 1-improper 2-colouring of CG,fori {1, 2}, the vertices a i and b i are coloured the same; and (iii) there exist two 1-improper 2-colourings c 1 and c 2 of CG such that c 1 (a 1 )= c 1 (b 1 )=c 1 (a 2 )=c 1 (b 2 )andc 2 (a 1 )=c 2 (b 1 ) c 2 (a 2 )=c 2 (b 2 ).

9 ABOUT A BROOKS-TYPE THEOREM 227 Properties (i) and (iii) can be directly checked. For property (ii), suppose that c is a 1-improper 2-colouring of CG. First, observe that, necessarily, three vertices among o 1,o 2,o 3, and o 4 are coloured the same, because the vertex o 0 is linked to all of them. Thus, the vertices o 2 and o 3 must be coloured differently. If c(a 1 ) c(b 1 ), then c(o 4 ) c(o 1 ) and by the preceding observation, the vertex o 0 has two neighbours of each colour, a contradiction. Now, suppose that c(a 2 ) c(b 2 )andsoc(o 3 ) c(o 7 ). By the construction, c(o 7 )=c(o 6 ) c(o 5 ), because o 6 and o 7 are joined by a regulator and o 5 and o 6 are joined by an inversor. Hence c(o 3 )=c(o 5 ), and so c(o 4 ) c(o 3 ). Therefore, c(o 4 )=c(o 2 )=c(o 1 ). This is not possible since c(o 4 )=c(o 7 ), so the vertex o 1 would have impropriety 2. a 2 o 3 o 2 o 4 b 1 o 1 o 0 o 5 a 1 o 7 o 6 b 2 Figure 7: The crossing gadget CG. A double edge stands for a regulator, and the bold edge stands for an inversor. It remains to show that Φ is satisfiable if and only if the obtained graph G Φ, which is planar and of maximum degree 4, is 1-improperly 2-colourable. Consider a 1-improper 2-colouring of G Φ. Without loss of generality, say that each clause vertices z of the clause gadgets is coloured 1. Then, at least one of the input vertices of each clause is coloured 1. Therefore, associating 1 with true and 2withfalse yields a truth assignment for Φ. Conversely, starting from a truth assignment of Φ, one can derive a 1-improper 2-colouring of G Φ as follows. Vertices corresponding to literals are coloured 1 if the corresponding literal is true, and2 otherwise. Thanks to the properties of the gadgets, such a partial colouring can be extended to a 1-improper 2-colouring of G Φ. Theorem 9. The following problem is N P -complete: INSTANCE: a planar graph G of maximum degree 6. QUESTION: is there a 2-improper 2-colouring of G?

10 228 R. CORRÊA, F. HAVET AND J-S. SERENI Proof. We shall reduce the problem to 1-improper 2-colouring of planar graphs with maximum degree 4. Let G be a planar graph of maximum degree 4: we construct, in polynomial time, a planar graph Ĝ of maximum degree 6 that is 2-improperly 2-colourable if and only if G is 1-improperly 2-colourable. The graph H showed in Figure 8 fulfils the following properties: (i) it is planar and has maximum degree 6; (ii) in any 2-improper 2-colouring of H, the vertex v must have impropriety at least 1; and (iii) there exists a 2-improper 2-colouring of H in which the vertex v has impropriety exactly 1. v a x b α β γ o y d e z Figure 8: The graph H. Property (i) can be directly checked. To prove (ii), it is sufficient to show that no 2-improper 2-colouring of H such that a and b are coloured the same exists. So, suppose that c(a) = c(b) = 1. If c(o) =1,thend, y, andx must be coloured 2, so e and z both receive colour 1. Therefore e has impropriety 3, because of b, o, andz, a contradiction. If c(o) = 2, then the three vertices α, β, γ cannot be coloured the same, otherwise b or o would have impropriety at least 3. Moreover, since c(b) =c(a) = 1, exactly two vertices among α, β, γ are coloured with colour 2. Hence, the vertices b and o both have impropriety 2 in the subgraph of G induced by the vertices a, b, o, α, β, γ. But the vertex e does not belong to this subgraph, and is linked to both b and o, a contradiction. This proves (ii). Assigning 1 to {v, a, d, e, α, β} and 2 to {b, o, x, y, z, γ}, we obtain the colouring of (iii). To construct the graph Ĝ, put a copy H(x) ofh for each vertex x V (G). Then, for each edge xy E(G), we put an edge between the vertex v of H(x) and the vertex v of H(y). Note that H has maximum degree 6 and v has degree 2 in H, so, as G has maximum degree 4, the graph Ĝ has maximum degree 6. Furthermore, the graph Ĝ is planar.

11 ABOUT A BROOKS-TYPE THEOREM 229 Now let c be a 1-improper 2-colouring of G. For any x V (G), assign the colour c(x) to the vertex v of H(x), and then extend the colouring to each copy of H by property (iii). If c is a 2-improper 2-colouring of Ĝ, thenforeachx V (G), assign to x the colour of the vertex v of H(x). The obtained 2-colouring of G is 1-improper because of property (ii) of H. References [1] K. Appel and W. Haken, Every planar map is four colorable I, Discharging, Illinois J. Math. 21(3) (1977), [2] K. Appel and W. Haken. Every planar map is four colorable, vol.98ofcontemporary Mathematics, Amer. Math. Soc., Providence, RI, 1989, (with the collaboration of J. Koch.) [3] K. Appel, W. Haken and J. Koch, Every planar map is four colorable II, Reducibility, Illinois J. Math. 21(3) (1977), [4] A. Beutelspacher and P.-R. Hering, Minimal graphs for which the chromatic number equals the maximal degree, Ars Combin. 18 (1984), [5] O. V. Borodin and A. V. Kostochka, On an upper bound of a graph s chromatic number, depending on the graph s degree and density, J. Combin. Theory Ser. B 23(2-3) (1977), [6] R. L. Brooks, On colouring the nodes of a network, Proc. Cambridge Philos. Soc. 37 (1941), [7] L. Cowen, R. H. Cowen and D. R. Woodall, Defective colorings of graphs in surfaces: partitions into subgraphs of bounded valency, J. Graph Theory 10(2) (1986), [8] L. Cowen, W. Goddard and C. E. Jesurum, Defective coloring revisited, J. Graph Theory 24(3) (1997), [9] N. Eaton and T. Hull, Defectivelist colorings of planar graphs, Bull. Inst. Combin. Appl. 25 (1999), [10] T. Emden-Weinert, S. Hougardy and B. Kreuter, Uniquely colourable graphs and the hardness of colouring graphs of large girth, Combin. Probab. Comput. 7(4) (1998), [11] B. Farzad, M. Molloy and B. Reed, (Δ k)-critical graphs, J. Combin. Theory Ser. B 93(2) (2005), [12] F. Havet, R. J. Kang and J.-S. Sereni, Improper colouring of unit disk graphs, (submitted). INRIA Research Report 6206, 2007.

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